# Tuesday February 18th :::{.theorem title="?"} Let $X/S$ be a projective subscheme (i.e. $X\subset \PP^n$ for some $n$). The Hilbert functor of flat families $\hilb_{X/S}^p$ is representable by a projective $S\dash$scheme. ::: :::{.remark} Note that without a fixed $P$, this is *locally* of finite type but not finite type. After fixing $P$, it becomes finite type. ::: :::{.example title="?"} For a curve of genus $g$, there is a smooth family $\mcc \mapsvia{\pi} S$ with $S$ finite-type over $\ZZ$ where every genus $g$ curve appears as a fiber. I.e., genus $g$ curves form a *bounded family* (here there are only finitely many algebraic parameters to specify a curve). How did we construct? Take the third power of the canonical bundle and show it's very ample, so it embeds into some projective space and has a Hilbert polynomial. ::: In fact, there is a finite type *moduli stack* $\mcm_g / \ZZ$ of genus $g$ curves. There will be a map $S \surjects \mcm_g$, noting that $\mcc$ is not a moduli space since it may have redundancy. We'll use the fact that a finite-type scheme surjects onto $\mcm_g$ to show it is finite type. :::{.remark} If $X/S$ is proper, we can't talk about the Hilbert polynomial, but the functor $\hilb_{X/S}$ is still representable by a locally finite-type scheme with connected components which are proper over $S$. ::: :::{.remark} If $X/S$ is *quasiprojective* (so locally closed, i.e. $X\injects \PP^n_S$), then $\hilb_{X/S}^P(T) \definedas \theset{z\in X_T \text{ projective, flat over S with fiberwise Hilbert polynomial P }}$ is still representable, but now by a quasiprojective scheme. ::: :::{.example title="?"} Length $Z$ subschemes of $\AA^1$: representable by $\AA^2$. ![Image](figures/2020-02-18-12:46.png)\ Upstairs: parametrizing length 1 subschemes, i.e. points. ::: :::{.remark} If $X\subset \PP_S^n$ and $E$ is a coherent sheaf on $X$, then \[ \Quot_{E, X/S}^{P}(T) = \theset{ j^*E \to F \to 0, \text{ over } X_T \to T,~F \text{ flat with fiberwise Hilbert polynomial } P } \] where $T \mapsvia{g} S$ is representable by an $S\dash$projective scheme. ::: :::{.example title="?"} Take $E = \OO_x$, $X$ and $S$ a point, and $E$ is a vector space, then $\Quot_{E/S}^P = \Gr(\rank, E)$. ::: :::{.warnings} The Hilbert scheme of 2 points on a surface is more complicated than just the symmetric product. ::: :::{.example title="?"} \[ \qty{\AA^2}^3 &\to \qty{\AA^2}^2 \\ \supseteq \Delta\definedas \Delta_{01} \cross \Delta_{02} &\to \qty{\AA^2}^2 \] where $\Delta_{ij}$ denote the diagonals on the $i, j$ factors. Here all associate points of $\Delta$ dominate the image, but it is not flat. Note that if we take the complement of the diagonal in the image, then the restriction $\Delta' \to \qty{\AA^2}^2\setminus D$ is in fact flat. ::: :::{.example title="Mumford"} The Hilbert scheme may have nontrivial scheme structure, i.e. this will be a "nice" Hilbert scheme with is generally not reduced. We will find a component $H$ of a $\hilb_{\PP^3_C}^P$ whose generic point corresponds to a smooth irreducible $C\subset \PP^3$ which is generically non-reduced. ::: ## Cubic Surfaces > See Hartshorne Chapter 5. Let $X\subset \PP^3$ be a smooth cubic surface, then $\OO(1)$ on $\PP^3$ restricts to a divisor class $H$ of a hyperplane section, i.e. the associated line bundle $\OO_x(H) = \OO_x(1)$. :::{.fact title="Important fact 1"} $X$ is the blowup of $\PP^2$ minus 6 points (replace each point with a curve). There is thus a blowdown map $X \mapsvia{\pi} \PP^2$. ![Image](figures/2020-02-18-13:07.png) Let $\ell = \pi^*(\text{line})$, then a fact is that $3\ell - E_1 -\cdots - E_6$ (where $E_i$ are the curves about the $p_i$) is very ample and embeds $X$ into $\PP^3$ as a cubic. ::: :::{.fact title="Important fact 2"} Every smooth cubic surface $X$ has *precisely* 27 lines. Any 6 pairwise skew lines arise as $E_1, \cdots, E_6$ as in the previous construction. ::: Take an $X$ and a line $L\subset X$. Consider any $C$ in the linear system $\abs{4H + 2L}$. Fact: $\OO(4H + 2L)$ is very ample, so embeds into a big projective space, and thus $C$ is smooth and irreducible by Bertini. Then the Hilbert polynomial of $C$ is of the form $at + b$ where $b = \chi(\OO_c)$, the Euler characteristic of the structure sheaf of $C$, and $a = \deg C$. So we'll compute these. We have $\deg C = H \cdot C$ (intersection) $= H \cdot(4H + 2L) = 4H^2 + 2H\cdot L = 4\cdot 3 + 2 = 14$. The intersections here correspond to taking hyperplane sections, intersecting with $X$ to get a curve, and counting intersection points: ![Image](figures/2020-02-18-13:14.png)\ In general, for $X$ a surface and $C\subset X$ a smooth curve, then $\omega_C = \omega_X(C)\mid_C$. Since $X\subset \PP^3$, we have \[ \omega_X &= \omega_{\PP^3}(X) \mid_X \\ &= \OO(-4) \oplus \OO(3)\mid_X \\ &= \OO_X(-1) \\ &= \OO_X(-H) .\] We also have \[ \omega_C &= \omega_X(C)\mid_X \\ &= \ro{ \qty{ \OO_X(-H) \oplus \OO_X(4H + 2L)}}{C} \\ \\ &\Downarrow \qquad \text{taking degrees} \\ \\ \deg \omega_C &= C\cdot(3H + 2L) \\ &= (4H+2L)(3H+2L) \\ &= 12H^2 + 14HL + 4L^2 \\ &= 36 + 14 + (-4) \\ &= 46 .\] Since this equals $2g(C) - 2$, we can conclude that the genus is given by $g(C) = 24$. Thus $P$ is given by $14t + (1-g) = 14t - 23$. :::{.remark} Good to know: moving a cubic surface moves the lines, you get a monodromy action, and the Weyl group of $E_6$ acts transitively so lines look the same. ::: :::{.claim title="1"} There is a flat family $Z\subset \PP^3_S$ with fiberwise Hilbert polynomial $P$ of cures of this form such that the image of the map $S \to \hilb_{\PP^3}^P$ has dimension 56. ::: :::{.proof title="of claim"} We can compute the dimension of the space of smooth cubic surfaces, since these live in $\PP H^0(\PP^3, \OO(3))$, which has dimension ${3+3\choose 3} -1 = 19$. Since there are 27 lines, the dimension of the space of such cubics with a choices of a line is also 19. Choose a general $C$ in the linear system $\abs{4H + 2L}$ will add $\dim \abs{4H + 2L} = \dim \PP H^0(x, \OO_x(C))$. We have an exact sequence \[ 0 \to \OO_X \to \OO_X(C) \to \OO_C(C) \to 0 \\ H^0\qty{ 0 \to \OO_X \to \OO_X(C) \to \OO_C(C) \to 0 } \\ .\] Since the first $H^0$ vanishes (?) we get an isomorphism. By Riemann-Roch, we have $$ \deg \OO_C(C) = C^2 = (4H+2L)^2 = 16H^2 + 16 HL + 4L^2 = 64 - 4 = 60 .$$ We can also compute $\chi(\OO_C(C)) = 60 - 23 = 37$. We have $$ h^0(\OO_C(C)) - h^1(\OO_C(C)) = h^0(\OO_C(C)) - h^0(\omega_C(-C))) = 2(23) - 60 < 0 ,$$ so there are no sections. So $\dim \abs{4H + 2L} = 37$. Thus letting $S$ be the space of cubic surfaces $X$, a line $L$, and a general $C \in \abs{4H + 2L}$, $\dim S = 56$. We get a map $S \to \hilb_{\PP^3}^P$, and we need to check that the fibers are 0-dimensional (so there are no redundancies). We then just need that every such $C$ lies on a unique cubic. Why does this have to be the case? If $C \subset X, X'$ then $C \subset X\intersect X'$ is degree 14 curve sitting inside a degree 6 curve, which can't happen. Thus if $H$ is a component of $\hilb_{\PP^3}^P$ containing the image of $S$, the $\dim H \geq 56$. ::: :::{.claim title="2"} For any $C$ above, we have $\dim T_C H = 57$. ::: When the subscheme is smooth, we have an identification with sections of the normal bundle $T_C H = H^0(C, N_{C/\PP^3})$. There's an exact sequence \[ 0 \to N_{C/X} = \OO_C(C) \to N_{C/\PP^3} \to N_{X/\PP^3}\mid_C = \OO_C(x)\mid_C = \OO_C(3H)\mid_C \to 0 .\] > Note $\omega_C = \OO_C(3H + 2L)$. As we computed, \[ H^0(\OO_C(C)) &= 37 \\ H^1(\OO_C(C)) &= 0 .\] So we need to understand the right-hand term $H^0(\OO_C(3H))$. By Serre duality, this equals $h^1(\omega_C(-3H)) = h^1(\OO_C(3L))$. We get an exact sequence \[ 0 \to \OO_X(2L-C) \to \OO_X(2L) \to \OO_C(2L) \to 0 .\] Taking homology, we have $0\to 0 \to 1 \to 1 \to 0$ since $2L-C = -4H$. Computing degrees yields $h^0 (\OO_C(3H)) = 20$. Thus the original exact sequence yields \[ 0 \to 37 \to ? \to 20 \to 0 ,\] so $? = 57$ and thus $\dim N_{C/\PP^3} = 57$. :::{.claim title="3"} \[ \dim H = 56 .\] ::: ### Proof That the Dimension is 56 Suppose otherwise. Then we have a family over $H^\mathrm{red}$ of *smooth* curves, where $f(S) \subset H^\mathrm{red}$, where the generic element is not on a cubic or any lower degree surface. Let $C'$ be a generic fiber. Then $C'$ lies on a pencil of quartics, i.e. 2 linearly independent quartics. Let $I = I_{C'}$ be the ideal of this curve in $\PP^3$, there is a SES $$ 0\to I(4) \to \OO(4) \to \OO_C(4) \to 0 .$$ It can be shown that $\dim H^0(I(4)) \geq 2$. :::{.fact} A generic quartic in this pencil is *smooth* (can be argued because of low degree and smoothness). ::: We can compute the dimension of quartics, which is ${4+3 \choose 3} - 1 = 35 - 1 = 34$. The dimension of $C'$s lying on a fixed quartic is $24$. But then the dimension of the image in the Hilbert scheme is at most $24 + 34 - 1 = 57$. It can be shown that the picard rank of such a quartic is 1, generated by $\OO(1)$, so this is a *strict* inequality, which is a contradiction since $\dim \hilb = 56$. This proves the theorem. :::{.remark} Use the fact that these curves are $K3$ surfaces? Get the fact about the generator of the Picard group from Hodge theory. So we can deform curves a bit, but not construct an algebraic family that escapes a particular cubic. :::